Genetics and Inheritance Worksheets - Free Printable

Let’s go step by step through each Punnett square problem. We’ll fill in the parents’ alleles, complete the squares, and calculate probabilities.

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Problem 1: Brown eyes (B) dominant over blue eyes (b)
Cross: Bb x bb

Parent 1 (Bb) can give B or b → top of square: B, b
Parent 2 (bb) can only give b → side of square: b, b

Punnett Square:

| | B | b |
|-----|-----|-----|
| b | Bb | bb |
| b | Bb | bb |

Genotypes:
- BB: 0 out of 4 → 0%
- Bb: 2 out of 4 → 50%
- bb: 2 out of 4 → 50%

✔ Probabilities:
- BB: 0
- Bb: 1/2 or 50%
- bb: 1/2 or 50%

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Problem 2: Curly hair (C) dominant over straight hair (c)
Cross: CC x Cc

Parent 1 (CC) gives only C → top: C, C
Parent 2 (Cc) gives C or c → side: C, c

Punnett Square:

| | C | C |
|-----|-----|-----|
| C | CC | CC |
| c | Cc | Cc |

Genotypes:
- CC: 2 out of 4 → 50%
- Cc: 2 out of 4 → 50%
- cc: 0 out of 4 → 0%

✔ Probabilities:
- CC: 1/2 or 50%
- Cc: 1/2 or 50%
- cc: 0

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Problem 3: Freckles (F) dominant over no freckles (f)

Part A: Homozygous recessive (ff) x Homozygous dominant (FF)

Parents: ff x FF
Parent 1 (ff) → f, f
Parent 2 (FF) → F, F

Punnett Square:

| | F | F |
|-----|-----|-----|
| f | Ff | Ff |
| f | Ff | Ff |

All offspring are Ff → all have freckles (since F is dominant)

✔ Probability:
- Freckles: 4/4 = 100% → 1
- No freckles: 0

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Part B: Heterozygous freckles (Ff) x Homozygous dominant freckles (FF)

Parents: Ff x FF
Parent 1 (Ff) → F, f
Parent 2 (FF) → F, F

Punnett Square:

| | F | F |
|-----|-----|-----|
| F | FF | FF |
| f | Ff | Ff |

Offspring: 2 FF, 2 Ff → ALL have at least one F → ALL have freckles

✔ Probability:
- Freckles: 4/4 = 100% → 1
- No freckles: 0

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Part C: Heterozygous freckles (Ff) x Heterozygous freckles (Ff)

Parents: Ff x Ff
Each parent gives F or f

Punnett Square:

| | F | f |
|-----|-----|-----|
| F | FF | Ff |
| f | Ff | ff |

Genotypes:
- FF: 1
- Ff: 2
- ff: 1

Phenotypes:
- Freckles: FF + Ff = 3 out of 4 → 75%
- No freckles: ff = 1 out of 4 → 25%

✔ Probability:
- Freckles: 3/4 or 75%
- No freckles: 1/4 or 25%

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Part D: Heterozygous freckles (Ff) x Homozygous recessive (ff)

Parents: Ff x ff
Parent 1 (Ff) → F, f
Parent 2 (ff) → f, f

Punnett Square:

| | F | f |
|-----|-----|-----|
| f | Ff | ff |
| f | Ff | ff |

Offspring:
- Ff: 2 → freckles
- ff: 2 → no freckles

✔ Probability:
- Freckles: 2/4 = 50% → 1/2
- No freckles: 2/4 = 50% → 1/2

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Final Answer:

Brown eyes (Bb x bb):
- BB: 0
- Bb: 1/2
- bb: 1/2

Curly hair (CC x Cc):
- CC: 1/2
- Cc: 1/2
- cc: 0

Freckles (ff x FF):
- Freckles: 1
- No freckles: 0

Freckles (Ff x FF):
- Freckles: 1
- No freckles: 0

Freckles (Ff x Ff):
- Freckles: 3/4
- No freckles: 1/4

Freckles (Ff x ff):
- Freckles: 1/2
- No freckles: 1/2